Problem 2

We consider a fixed point

P

in the interior of a fixed sphere. We construct three segments

PA, PB, PC

, perpendicular two by two, with the vertices

A, B, C

on the sphere. We consider the vertex

Q

which is opposite to

P

in the parallelepiped (with right angles) with

PA, PB, PC

as edges. Find the locus of the point

Q

when

A, B, C

take all the positions compatible with our problem.

Step 7 of 7: Conclude: the locus is a sphere centered at O

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In plain words

A constant distance from a fixed center is exactly the defining property of a sphere.

OQ = \sqrt{3R^2-2d^2} \text{ (constant)}

Detailed analysis

Since $|OQ|=\sqrt{3R^2-2d^2}$ is a fixed constant independent of $A,B,C$ (and this value is real because $d<R$ ), the locus of $Q$ as $A,B,C$ range over all valid orthogonal triples is exactly the sphere centered at $O$ with this radius.